Is there variation of the binomial theorem as follows?
$$\sum_{i=0}^n {m \choose i} a^{m-i} b^i $$
I am trying to find a formula for it that is a function of $n$ without summation notation. I think you would call this is a solution in closed form. Any help would be great. Thanks.
This may not be what you were hoping for but we certainly can write your sum as a function of $n$ without summation notation using this expression for truncated hypergeometric series. We have $$ \sum_{i=0}^n\binom{m}{i}a^{m-i}b^i=a^m(-b/a)^n\frac{(-m)_n}{n!}{_2F}_1\left({-n,1 \atop 1-n+m};-\frac{a}{b}\right), $$ where $(x)_n$ is the Pochhammer symbol. Using transformations for the hypergeometric function may be able to put this in a more compact form.